Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int par[17][100005], d[100005];
vector<int> adj[100005];
struct node {
  int a[11];
  node() { memset(a, 63, sizeof(a)); }
  void insert(int j) {
    a[10] = j;
    sort(a, a + 11);
  }
} vals[17][100005];
node merge(node a, node b) {
  node temp = a;
  for (int i = 0; i < 11; i++) {
    temp.insert(b.a[i]);
  }
  return temp;
}
void dfs(int p, int v) {
  par[0][v] = p;
  for (int i = 1; i < 17; i++) {
    par[i][v] = par[i - 1][par[i - 1][v]];
    vals[i][v] = merge(vals[i - 1][v], vals[i - 1][par[i - 1][v]]);
  }
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != p) {
      d[u] = d[v] + 1;
      dfs(v, u);
    }
  }
}
int get_parent(int v, int k) {
  for (int i = 0; i < 17; i++) {
    if ((1 << i) & k) {
      v = par[i][v];
    }
  }
  return v;
}
int lca(int u, int v) {
  if (d[u] < d[v]) {
    swap(u, v);
  }
  u = get_parent(u, d[u] - d[v]);
  if (u == v) {
    return u;
  }
  for (int i = 17 - 1; i >= 0; i--) {
    if (par[i][u] != par[i][v]) {
      v = par[i][v];
      u = par[i][u];
    }
  }
  return par[0][v];
}
node get(int v, int k) {
  node ans;
  for (int i = 0; i < 17; i++) {
    if ((1 << i) & k) {
      ans = merge(ans, vals[i][v]);
      v = par[i][v];
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, m, q;
  cin >> n;
  cin >> m;
  cin >> q;
  for (int x = 0; x < n - 1; x++) {
    int v, u;
    cin >> v;
    cin >> u;
    u--;
    v--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  for (int i = 0; i < m; i++) {
    int c;
    cin >> c;
    c--;
    vals[0][c].insert(i);
  }
  dfs(0, 0);
  while (q--) {
    int u, v, k;
    cin >> u >> v >> k;
    u--;
    v--;
    int p = lca(u, v);
    node x = get(u, d[u] - d[p]);
    node y = get(v, d[v] - d[p] + 1);
    node ans = merge(x, y);
    int tmp = 0;
    while (tmp < k && ans.a[tmp] < m) tmp++;
    k = tmp;
    cout << k;
    for (int i = 0; i < k; i++) cout << " " << ans.a[i] + 1;
    cout << "\n";
  }
  return 0;
}